The maximum absolute error correct to the nearest $10\text{ g}$
$\begin{array}{cl}
= & \dfrac{1}{2} \times 10 \\
= & 5\text{ g}
\end{array}$
Therefore, the minimum possible weight of a regular packet of cheese
$\begin{array}{cl}
= & 220 -5 \\
= & 215\text{ g}
\end{array}$
Therefore, the minimum possible weight of $250$ regular packets of cheese
$\begin{array}{cl}
= & 215 \times 250 \\
= & 53750 \text{ g} \\
= & 53.75\text{ kg}
\end{array}$
The maximum absolute error correct to the nearest $0.1\text{ kg}$
$\begin{array}{cl}
= & \dfrac{1}{2} \times 0.1 \\
= & 0.05\text{ kg}
\end{array}$
Therefore, the upper limit of $53.6\text{ kg}$
$\begin{array}{cl}
= & 53.6 + 0.05 \\
= & 53.65 \text{ kg}
\end{array}$
Since the minimum possible weight of $250$ regular packets of cheese is greater than the upper limit of $53.6\text{ kg}$, then the claim is not correct.
